SecondaryStorage
Question 1 
Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is _________.
14020  
14021  
14022  
14023 
Question 1 Explanation:
Given
Seek time = 4ms
60s→ 10000 rotations
∴ Rotational latency =1/2×6ms=3ms
1sector→600sectors
⇒6ms←600sectors (1 rotation means 600 sectors (or)1 track)
1 track→6ms/600=0.01ms
2000sector→2000(0.01)=20ms
∴total time needed to read the entire file is
=2000(4+3)+20
=8000+6000+20
=14020ms
Seek time = 4ms
60s→ 10000 rotations
∴ Rotational latency =1/2×6ms=3ms
1sector→600sectors
⇒6ms←600sectors (1 rotation means 600 sectors (or)1 track)
1 track→6ms/600=0.01ms
2000sector→2000(0.01)=20ms
∴total time needed to read the entire file is
=2000(4+3)+20
=8000+6000+20
=14020ms
Question 2 
Consider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 50×10^{6} bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller’s transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512byte sector of the disk is _____________.
6  
6.1  
6.2  
6.3 
Question 2 Explanation:
15000 rotations → 60 sec
1 rotation → 60 /15000 sec = 4 ms
∴ Average rotational delay = 1/2 × 4 ms = 2 ms
Average seek time = 2 × Average rotational delay
= 2 × 2 ms = 4 ms
Time to transfer 512 byte,
50 × 10^{6}B  1s
512 B  512B/ 50×10^{6}B/s = 0.01 ms
∴ Controllers transfer time
= 10 × Disk transfer time
= 10 × 0.01 ms
= 0.01 ms
∴ Avg. time = (4 + 2 + 0.1 + 0.01)ms
= 6.11 ms
1 rotation → 60 /15000 sec = 4 ms
∴ Average rotational delay = 1/2 × 4 ms = 2 ms
Average seek time = 2 × Average rotational delay
= 2 × 2 ms = 4 ms
Time to transfer 512 byte,
50 × 10^{6}B  1s
512 B  512B/ 50×10^{6}B/s = 0.01 ms
∴ Controllers transfer time
= 10 × Disk transfer time
= 10 × 0.01 ms
= 0.01 ms
∴ Avg. time = (4 + 2 + 0.1 + 0.01)ms
= 6.11 ms
Question 3 
An application loads 100 libraries at startup. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10 ms. Rotational speed of disk is 6000 rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected).
0.50 s  
1.50 s  
1.25 s  
1.00 s 
Question 3 Explanation:
Transfer Time = Seek Time + Rotational latency/2 + time to read/write
= 10ms + (60/12000)+0
For 100 libraries = 100(10ms + (60/12000)+0) = 1.5seconds
= 10ms + (60/12000)+0
For 100 libraries = 100(10ms + (60/12000)+0) = 1.5seconds
Question 4 
A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple c, h, s , where <c,h,s>, where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0^{th} sector is addressed as 0, 0, 0 , the 1^{st} sector as 0, 0,1 , and so on
The address <400, 16, 29> corre4sponds tp sector number:
505035  
505036  
505037  
505038 
Question 4 Explanation:
By default the cylinders numbering starts from 0. In the given hard disk there are 1000 cylinders, 1st cylinder is numbered 0, 2nd cylinder is numbered 1 and so on 1000th cylinder is numbered 999.
To reach a cylinder numbered n, we need to cross cylinders numbered from 0 to n1.
To reach cylinder numbered 400 (401th cylinder) we need to skip cylinders numbered from 0 to 399, (from cylinder number 0 to 399 there are total 400 cylinders).
Each cylinder consists of 10 plates with 2 recording surfaces and 63 sectors per track.
So total number of tracks to be crossed to reach cylinder numbered 400 = 400 * (10*2) * 63 = 504,000 sectors.
Then, to reach the 16th surface of the cylinder numbered 400, we need to skip another 16*63 = 1,008 sectors.
Finally, to find the 29 sector, we need to move another 29 sectors.
In total, we moved 504,000 + 1,008 + 29 = 505,037 sectors.
Hence, option C is the answer.
To reach a cylinder numbered n, we need to cross cylinders numbered from 0 to n1.
To reach cylinder numbered 400 (401th cylinder) we need to skip cylinders numbered from 0 to 399, (from cylinder number 0 to 399 there are total 400 cylinders).
Each cylinder consists of 10 plates with 2 recording surfaces and 63 sectors per track.
So total number of tracks to be crossed to reach cylinder numbered 400 = 400 * (10*2) * 63 = 504,000 sectors.
Then, to reach the 16th surface of the cylinder numbered 400, we need to skip another 16*63 = 1,008 sectors.
Finally, to find the 29 sector, we need to move another 29 sectors.
In total, we moved 504,000 + 1,008 + 29 = 505,037 sectors.
Hence, option C is the answer.
Question 5 
〈0, 15, 31〉
 
〈0, 16, 30〉  
〈0, 16, 31〉
 
〈0, 17, 31〉 
Question 5 Explanation:
We know in each track there are 63 sectors. And we know there is one track per surface.
From the given options we can calculate the sector numbers as
Option A  15*63+31 = 976
Option B  16*63+30 = 1038
Option C  16*63+31 = 1039
Option D  17*63+31 = 1102
Hence Option C is the answer.
From the given options we can calculate the sector numbers as
Option A  15*63+31 = 976
Option B  16*63+30 = 1038
Option C  16*63+31 = 1039
Option D  17*63+31 = 1102
Hence Option C is the answer.
Question 6 
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to
nonuniform distribution of requests
 
arm starting and stopping inertia  
higher capacity of tracks on the periphery of the platter  
use of unfair arm scheduling policies

Question 6 Explanation:
Whenever the head moves from one track to another track its speed changes, this is the case of inertia .
Because the definition of inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Because the definition of inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Question 7 
Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:
256 Mbyte, 19 bits
 
256 Mbyte, 28 bits
 
512 Mbyte, 20 bits  
64 Gbyte, 28 bit

Question 7 Explanation:
Given that the disk pack has 16 surfaces, 128 tracks per surface, 256 sectors per track and each sector size is 512 bytes.
So the disk pack capacity = 16 * 128 * 256 * 512 bytes = 256 MB
To specify a sector we need the information about surface number, track number and sector number within a track. Surface number needs 4 bits as there are 16 surfaces(2^{4}), track number needs 7 bits as there are 128 tracks(2^{7}) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (2^{8}). Total number bits needed to specify a particular sector = 4+7+8 = 19 bits. Hence option A is the answer.
So the disk pack capacity = 16 * 128 * 256 * 512 bytes = 256 MB
To specify a sector we need the information about surface number, track number and sector number within a track. Surface number needs 4 bits as there are 16 surfaces(2^{4}), track number needs 7 bits as there are 128 tracks(2^{7}) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (2^{8}). Total number bits needed to specify a particular sector = 4+7+8 = 19 bits. Hence option A is the answer.
Question 8 
A hard disk system has the following parameters :

 Number of tracks = 500
 Number of sectors/track = 100
 Number of bytes /sector = 500
 Time taken by the head to move from one track to adjacent track = 1 ms
 Rotation speed = 600 rpm.
300.5 ms  
255.5 ms  
255.0 ms  
300.0 ms 
Question 8 Explanation:
Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time
For Avg. seek time:
Given that, time to move between successive tracks is 1ms.
Time to move from track1 to track1 = 0ms
Time to move from track1 to track2 = 1ms
Time to move from track1 to track3 = 2ms
⋮ Time to move from track1 to track500 = 499ms
∴ Avg. seek time = 0+1+2+...+499/500 = 249.5ms
Avg. rotational delay:
600 rotations in 60sec.
One rotation takes 60/600 s = 100ms ∴ Avg. rotational delay = 100/2 = 50ms
Data transfer time:
In one rotation, we can read data on one complete track
= 100 × 500 = 50,000B data is read in one complete rotation
One complete rotation takes 100ms.
100ms → 50,000B
250B → 100/50000 × 250 = 0.5ms
∴ Avg. time to transfer = 249.5 × 50 + 0.5 = 300ms
For Avg. seek time:
Given that, time to move between successive tracks is 1ms.
Time to move from track1 to track1 = 0ms
Time to move from track1 to track2 = 1ms
Time to move from track1 to track3 = 2ms
⋮ Time to move from track1 to track500 = 499ms
∴ Avg. seek time = 0+1+2+...+499/500 = 249.5ms
Avg. rotational delay:
600 rotations in 60sec.
One rotation takes 60/600 s = 100ms ∴ Avg. rotational delay = 100/2 = 50ms
Data transfer time:
In one rotation, we can read data on one complete track
= 100 × 500 = 50,000B data is read in one complete rotation
One complete rotation takes 100ms.
100ms → 50,000B
250B → 100/50000 × 250 = 0.5ms
∴ Avg. time to transfer = 249.5 × 50 + 0.5 = 300ms
Question 9 
Consider a disk drive with the following specifications: 16 surfaces, 512 tracks/surface, 512 sectors/track, 1 KB/sector, rotation speed 3000 rpm. The disk is operated in cycle stealing mode whereby whenever one byte word is ready it is sent to memory; similarly, for writing, the disk interface reads a 4 byte word from the memory in each DMA cycle. Memory cycle time is 40 nsec. The maximum percentage of time that the CPU gets blocked during DMA operation is:
10  
25  
40  
50 
Question 9 Explanation:
Time for one rotation = 60/3000
It reads 512 * 1024 B in one rotation.
Time taken to read 4B = 153ns
(153) for 4 is approximately = 160
Percentage CPU get blocked = 40*100/160 = 25
It reads 512 * 1024 B in one rotation.
Time taken to read 4B = 153ns
(153) for 4 is approximately = 160
Percentage CPU get blocked = 40*100/160 = 25
Question 10 
What is the swap space in the disk used for?
Saving temporary html pages
 
Saving process data
 
Storing the superblock
 
Storing device drivers

Question 10 Explanation:
A swap file is a space on hard disk used as the virtual memory extension of a computer's real memory.
→ The interchange of data between a virtual memory and a real memory is called as swapping and space on a disk as 'swap space'.
→ Swap space in the disk can be used for saving process data.
→ The interchange of data between a virtual memory and a real memory is called as swapping and space on a disk as 'swap space'.
→ Swap space in the disk can be used for saving process data.
There are 12 questions to complete.